Question: $h(x) = 17+\dfrac{x}{6}$ $h(-18)=$
To find the value of $h({-18})$, we need to substitute ${x}={-18}$ into the function's formula: $\begin{aligned}h({x})&=17+\dfrac{{x}}{6}\\\\ h({-18})&=17+\dfrac{{-18}}{6}\\\\ &=17-3\\\\ &=14\end{aligned}$ In conclusion, $h(-18)=14$